A detector at `x=0` receives waves from three sources each of amplitude `A` and frequencies `f + 2,f` and `f-2`.
The time at which intensity is maximum, is

Options

A. `t=0,1//4,1//2,3//4,…sec`
B. `t=1//6,1//3,2//3,5//6,…sec`
C. `t=0,1//2,3//2,5//2,..sec`
D. `t=1//2,1//4,1//6,1//8,…sec`

Explanation

To find the time at which the intensity is maximum at the detector located at \( x = 0 \), we can follow these steps:

### Step 1: Understand the wave sources
We have three sources emitting waves with the following frequencies:
- Source 1: \( f + 2 \)
- Source 2: \( f \)
- Source 3: \( f - 2 \)

### Step 2: Write the wave equations
The wave equations for the three sources can be expressed as:
- \( y_1(t) = A \sin(2\pi (f + 2)t) \)
- \( y_2(t) = A \sin(2\pi ft) \)
- \( y_3(t) = A \sin(2\pi (f - 2)t) \)

### Step 3: Combine the waves
The total wave at the detector is the sum of the individual waves:
\[ y(t) = y_1(t) + y_2(t) + y_3(t) \]
This can be simplified using the sine addition formula.

### Step 4: Find the resultant frequency
The frequencies of the waves are \( f + 2 \), \( f \), and \( f - 2 \). The maximum frequency difference is \( 4 \) (from \( f + 2 \) to \( f - 2 \)). The resultant frequency of the combined wave will be the average of these frequencies:
\[ f_{avg} = f \]
The beat frequency will be \( 4 \) (the difference between the highest and lowest frequencies).

### Step 5: Determine the condition for maximum intensity
The intensity of the resultant wave is maximum when the phase difference between the waves is an integer multiple of \( 2\pi \). This occurs when:
\[ \sin(4\pi t) = 0 \]
This means:
\[ 4\pi t = n\pi \quad (n \text{ is an integer}) \]
Thus:
\[ t = \frac{n}{4} \]

### Step 6: Identify specific times
The times at which the intensity is maximum are:
- \( t = 0 \)
- \( t = \frac{1}{4} \)
- \( t = \frac{1}{2} \)
- \( t = \frac{3}{4} \)
- and so on...

### Conclusion
The times at which the intensity is maximum are given by:
\[ t = \frac{n}{4} \quad (n = 0, 1, 2, 3, \ldots) \]